4 Answers
4

Think of an element of $\Bbb Z_n \times \Bbb Z_m$ as having two parts, written $\langle p,q\rangle$. The left part $p$ is an element of $\Bbb Z_n$ and the right part $q$ is an element of $\Bbb Z_m$. The two parts never interact. To add $\langle a, b\rangle$ to
$\langle c, d\rangle$, you add the two parts separately, and get
$\langle a+c, b+d\rangle$. The left-side addition is done $\Bbb Z_n$-style, because $a$ and $c$ are elements of $\Bbb Z_n$, and the right-side addition is done $\Bbb Z_m$-style.

But $\Bbb Z_4$ is different; it has two elements, 1 and 3, that do not have the property that $x+x=0$. They have $1+1 = 3+3 = 2$ instead, and in $\Bbb Z_4, 2\ne 0$. $\V$ has nothing like this.

Or looked at in the opposite direction, $\Bbb Z_4$ has an operation, namely the operation of adding 1, which you must do four times before you get back to where you started; it is analogous to giving an object a quarter turn. After four quarter turns, and no fewer, the object has returned to its original position. $\V$ has nothing like this; every operation in $\V$ gets you back to where you started after at no more than two repetitions.

$\mathbb Z_2\times\mathbb Z_2$ is the set of ordered pairs $(a,b)$ with $a,b\in\mathbb Z_2$, i.e. $\{(0,0),(1,0),(0,1),(1,1)\}$. The group operation is coordinate-wise $\mathbb Z_2$-addition.

It cannot be isomorphic to $\mathbb Z_4$ because all nonzero elements in $\mathbb Z_2\times\mathbb Z_2$ have order 2, while in $\mathbb Z_4$ there are elements of order 4. In particular, $\mathbb Z_2\times\mathbb Z_2$ is not cyclic.

(I'm omitting in both cases to put some $\overline{2}$ all above those numbers in order to point out that they are not actual natural numbers, but classes. But this way I can write faster.)

So, this group $\mathbb{Z}_4$ has a particular guy, $1$, inside with this property:

$$
1 + 1 = 2, \ 1 + 1 + 1 = 3, \ 1 + 1 + 1 + 1 = 4 = 0 \ .
$$

Can you find anyone inside $\mathbb{Z}_2\times \mathbb{Z}_2$ with a similar behaviour? Namely, that four times itself produces the neutral element? [EDIT: Thanks to MJD's remark. And not just two times?] Nope? You can't? -Then, they cannot be isomorphic.